Answer:
Q=0.95 W/m
Explanation:
Given that
Outer diameter = 0.3 m
Thermal conductivity of material
[tex]K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}[/tex]
So the mean conductivity
[tex]K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )[/tex]
[tex]T_m=\dfrac{160+273+40+273}{2}[/tex]
[tex]T_m=373 K[/tex]
[tex]K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )[/tex]
[tex]K_m=0.112 \frac{W}{mK}[/tex]
So heat conduction through cylinder
[tex]Q=kA\dfrac{\Delta T}{L}[/tex]
[tex]Q=0.112\times \pi \times 0.15^2\times 120[/tex]
Q=0.95 W/m